Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

walters Group Title

Question based on (Groups)

  • one year ago
  • one year ago

  • This Question is Closed
  1. walters Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    jjj let the mapping \[\tau _{ab}\] for a,b element R ,maps the reals by the ruel \[\tau _{ab}:x \rightarrow ax+b.Let G={\tau _{ab}:a \neq 0}\] Determine whether or not G is an abelian group under the composition of mappings

    • one year ago
  2. walters Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    oops ignore jjj

    • one year ago
  3. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    Well, let \(\tau_{ab},\tau_{cd}\in G\). Then \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b.\]Also, \[\tau_{cd}\circ\tau_{ab}(x)=\tau_{cd}(ax+b)=cax+bc+d.\]Since \(acx+ad+b\neq acx+bc+d\), it must be that \(G\) is not abelian.

    • one year ago
  4. walters Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    am i not suppose to check whether is a group using the condition before i can conclude that is an abelian group

    • one year ago
  5. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    It just asked if it was an abelian group. Since the elements don't commute, it can't be an abelian group even if it is a group. However, if we want to check it's a group, we can do that as well. Have you shown any of the axioms for proving it's a group yet?

    • one year ago
  6. walters Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i am failing to show the axioms

    • one year ago
  7. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    First, we can see that \(G\) is closed under composition. \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b=\tau_{(ac)(ad+b)}.\]We also have the identity \(\tau_{10}\). Now we just have to check associativity and inverses.

    • one year ago
  8. walters Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    why do u chose to use |dw:1360875075213:dw|

    • one year ago
  9. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    "\(\circ\)" is a relatively common notation to denote the composition of two functions. You can also use \(\tau_{ab}(\tau_{cd}(x))\) is you prefer.

    • one year ago
  10. walters Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    can u also use * if u wnt

    • one year ago
  11. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    For inverses, we want \[\tau_{ab}\circ\tau_{cd}(x)=acx+ad+b=x\]for some \(c,d\in\mathbb{R}\). We can see that \(c=a^{-1}\) and \(d=a^{-1}(-b)\) work for this direction. For the other, \[\tau_{cd}\circ\tau_{ab}(x)=cax+bc+d=a^{-1}ax+ba^{-1}-a^{-1}b=x,\]so we do indeed have inverses.

    • one year ago
  12. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    If you make it clear your group function is composition of functions, you can use * as well.

    • one year ago
  13. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    As for associativity, I'm just going to use the fact that composition of functions is always associative. If you have not heard of that yet, we can still go over how to prove this particular function is associative.

    • one year ago
  14. walters Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    please do

    • one year ago
  15. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    let \(\tau_{ab},\tau_{cd},\tau_{ef}\in G\). Then\[\tau_{ab}\circ(\tau_{cd}\circ\tau_{ef})(x)=\tau_{ab}=\tau_{ab}(cex+cf+d)=acex+acf+ad+b\]and\[(\tau_{ab}\circ\tau_{cd})\circ\tau_{ef}(x)=(\tau_{(ac)(ad+b)})(ex+f)=acex+acf+ad+b\]Since these are equal, it's associative, and we have a group.

    • one year ago
  16. walters Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    k it is a group but not an abelian group

    • one year ago
  17. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    Yup.

    • one year ago
  18. walters Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    thnx

    • one year ago
  19. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    You're welcome.

    • one year ago
  20. walters Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    can we verify identity

    • one year ago
  21. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 4

    \[\tau_{10}(x)=1x+0=x\]

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.